# Thread: Can I get some help solving logerithms?

1. ## Can I get some help solving logerithms?

log10 SQ of 10

I did

10^y=SQ of 10

Y=1/2 but I don't know how to that. I know I have to do something with the SQ of 10. It's not a perfect square. it comes out to 3.19.

Thanks

Jason

2. Originally Posted by Darkhrse99
log10 SQ of 10

I did

10^y=SQ of 10

Y=1/2 but I don't know how to that. I know I have to do something with the SQ of 10. It's not a perfect square. it comes out to 3.19.

Thanks

Jason
$log_{10}(\sqrt{10}) = log_{10}(10^{1/2}) = \frac{1}{2}log_{10}(10) = \frac{1}{2} \cdot 1 = \frac{1}{2}$

-Dan

3. Dan

How did you get (10)^1/2 out of SQRT of 10?

4. Originally Posted by Darkhrse99
Dan

How did you get (10)^1/2 out of SQRT of 10?
For any $a > 0$
$\sqrt[n]{a} = a^{1/n}$

We can prove this for the square roots fairly easily by solving $x^2 = a$ in two ways:
$x^2 = a$

$x = \sqrt{a}$ <-- Taking just the principle value for the moment.

And
$x^2 = a$

$(x^2)^{1/2} = a^{1/2}$

$x = a^{1/2}$

Thus
$\sqrt{a} = a^{1/2}$.

-Dan

5. Thanks Dan.

6. Originally Posted by Darkhrse99
Thanks Dan.
I live to serve.

You are welcome!

-Dan